2.5. Algebraic Identities

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it.

An algebraic identity is a statement that equates two algebraic expressions and holds true for every possible value of the variables in these expressions. This means that if you substitute any value into the identity, both sides will give the same result. For example, the identity (x + y)² = x² + 2xy + y² works for any numbers you choose for x and y.

Think of an algebraic identity like a recipe that works every time. Just like you can substitute different ingredients and still produce a delicious cake, you can change the values of the variables in an identity and it will still hold true, just like the recipe's instructions produce the same result.

You have studied the following algebraic identities in earlier classes:

Identity I : (x + y)² = x² + 2xy + y² Identity II : (x – y)² = x² – 2xy + y² Identity III : x² – y² = (x + y)(x – y) Identity IV : (x + a)(x + b) = x² + (a + b)x + ab.

Here are some common algebraic identities:

1. Identity I states the expansion of the square of a sum.
2. Identity II is the expansion of the square of a difference.
3. Identity III shows the difference of squares and how it can be factored into two binomials.
4. Identity IV allows us to expand the product of two binomials. These identities are foundational tools in algebra that help simplify expressions and solve equations.

Consider these identities like shortcuts in a manual. Instead of going through all the steps to get to a final product (the solution), these identities help you quickly get there without getting lost in unnecessary calculations. They are like knowing a direct route in a city that saves time.

Example 1: Find the following products using appropriate identities: (i) (x + 3)(x + 3) (ii) (x – 3)(x + 5)

Solution: (i) Here we can use Identity I: (x + y)² = x² + 2xy + y². Putting y = 3 in it, (x + 3)(x + 3) = (x + 3)² = x² + 2(x)(3) + (3)² = x² + 6x + 9 (ii) Using Identity IV, we have (x – 3)(x + 5) = x² + (–3 + 5)x + (–3)(5) = x² + 2x – 15.

In these examples, we see how to apply the identities to simplify products:

1. For (x + 3)(x + 3), we recognize it as (x + 3)² and use Identity I to expand it.
2. For (x – 3)(x + 5), we utilize Identity IV to expand by substituting the appropriate values. This saves a lot of time and effort versus multiplying out each term.

Imagine you are packing boxes. Instead of figuring out how many boxes you need for every single item, you use the identities like a packing formula that quickly tells you how many full boxes can be packed together. This not only makes the task faster but also ensures accuracy in your packing.

You have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as you can see in the following examples. Example 13: Factorise: (i) 49a² + 70ab + 25b² (ii) 4y² – 4y + 1.

In these examples, we apply our identities to factor algebraic expressions:

1. For 49a² + 70ab + 25b², we recognize it as the expanded form of Identity I. We find that it factors to (7a + 5b)².
2. For 4y² – 4y + 1, we can see this fits another identity and factors to (2y – 1)². These steps show how identities can help re-write expressions in a simpler form.

Consider factoring algebraic expressions like peeling an onion to get to the core. Just as you carefully peel off layers to reach the center, using identities allows you to simplify complex expressions from their outer layer back to their basic components.

So far, all our identities involved products of binomials. Let us now extend the Identity I to a trinomial x + y + z.

We can extend our understanding of identities by exploring trinomials. For example, when computing (x + y + z)², we use the idea that we can first combine x and y, then add z to explore the expansion systematically. This allows us to create a new identity: Identity V: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Think of this process as organizing a group project. At first, you might group team members to tackle individual tasks. Once each pair works, you combine results to get the final output, similar to how we combine different terms to reach a more comprehensive identity when dealing with three variables.

Example 14: Write (3a + 4b + 5c)² in expanded form. Solution: Using Identity V, we have (3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a).

Example 15: Expand (4a – 2b – 3c)².

In these examples, we see how to apply the trinomial identity to obtain expanded forms. For (3a + 4b + 5c)², we follow Identity V to expand it to 9a² + 16b² + 25c² + 24ab + 40bc + 30ac. Similarly, (4a – 2b – 3c)² also applies Identity V correctly by substituting negative values.

Imagine planning a big event. You can think of the expanded form like organizing food, venues, decorations, and music into sections to ensure that every detail is covered, succinctly bringing together every part of the event into a coherent plan, just like the terms in the expanded form.

So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute (x + y)³.

We start by using the square expansion and then multiplying it by (x + y) to extend into cubic terms. This leads us to Identity VI: (x + y)³ = x³ + y³ + 3xy(x + y). We can also see how applying the negative of y gives us Identity VII: (x – y)³.

Think of cubic identities like building a three-dimensional space. Just as every room in a house has dimensions that relate to each other, cubic identities relate the variables in a point-by-point expansion, demonstrating how different factors interact in three dimensions.

Example 17: Write the following cubes in the expanded form: (i) (3a + 4b)³ (ii) (5p – 3q)³.

In these examples, we leverage cubic identities for efficient calculations. For (3a + 4b)³, we identify x = 3a and y = 4b, using Identity VI to expand it to 27a³ + 64b³ + 108a²b + 144ab². For (5p – 3q)³, we apply Identity VII for a similar expansion approach.

Imagine packing three-dimensional boxes with various items. Writing out the expanded form is like noting down the capacity and arrangement of these boxes, ensuring all possible combinations are counted, similar to how we expand variables in algebraic forms.

Example 19: Factorise 8x³ + 27y³ + 36x²y + 54xy².

Solution: (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)² = (2x + 3y)³.

Here, we recognize the structure of the cubic identity in evaluating the factorization. By rearranging and applying the identities accordingly, we simplify a complex polynomial into a neat cubic factor.

Think of evaluating and factorizing using identities as organizing your closet. When everything is neatly categorized, finding an outfit becomes simple. Similarly, using identities neatly organizes the polynomial, making it easier to factor and reveal its foundational structure.

You have learned various identities that can help you not only solve equations but also understand how polynomials interact through their structures.

The study of algebraic identities unlocks deeper comprehension in mathematics, helping us manipulate polynomial expressions effectively. By recognizing patterns and applying these identities, we can simplify and solve problems more efficiently.

Understanding these identities is like learning a new language. Once you know the keywords and grammar (the respective identities), you can communicate more effectively and understand conversations (equations and expressions) with greater clarity.

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